Palindrome partitioning [leetcode] DFS and DP Solutions

The first method is DFS, which finds all possible prefixes and recursively calls partition (remaining string)

Complexity: O (2 ^ N)

The Code is as follows:

    vector<vector<string>> partition(string s) {        vector<vector<string>> res;        vector<string> patition;        if (s.size() == 0) return res;        partition(s, patition, res);        return res;    }        void partition(string s, vector<string>& patition, vector<vector<string>>& res) {        for (int i = 1; i < s.size(); i++)        {            if (isPalindrome(s.substr(0, i)))            {                patition.push_back(s.substr(0, i));                partition(s.substr(i), patition, res);                patition.pop_back();            }        }        if (isPalindrome(s))        {            patition.push_back(s);            res.push_back(patition);            patition.pop_back();        }    }        bool isPalindrome(string s)    {        int l = 0, r = s.size() - 1;        while (l <= r)        {            if (s[l] != s[r]) return false;            l++;            r--;        }        return true;    }

The second method is DP.

From https://oj.leetcode.com/discuss/9623/my-java-dp-only-solution-without-recursion-o-n-2

Res (I) represents all decomposition methods for S [0... I-1]

Ispalin (I, j) indicates whether s [I... J] is a return string.

Ispalin (I, j) = true if I = J or (s [I] = s [J] And ispalin (I + 1, J-1 )) or (s [I] = s [J] And I + 1 = J)

Res (I) = res (j) + s [j... I-1] If ispalin (J, I-1)

    vector<vector<string>> partition(string s) {        int size = s.size();        vector<vector<vector<string>>> res(size + 1);        res[0].push_back(vector<string>(0));        vector<vector<bool>> isPalin(size + 1, vector<bool>(size + 1, false));                for (int i = 0; i < size; i++)        {            for (int j = 0; j <= i; j++)            {                if (i == j || s[i] == s[j] && (j + 1 == i || isPalin[j + 1][i - 1]))                {                    isPalin[j][i] = true;                    for (int p = 0; p < res[j].size(); p++)                    {                        vector<string> prefix = res[j][p];                        prefix.push_back(s.substr(j, i - j + 1));                        res[i + 1].push_back(prefix);                    }                }            }        }        return res[size];    }

Palindrome partitioning [leetcode] DFS and DP Solutions